Doug Goncz
May12-04, 05:21 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hey, gang.\n\nI\'ve completed fourth order Runge-Kutta simulation of electrically augmented,\nflywheel augmented, and standard bicycles on USGS terrain routes in my\nneighborhood. The results are that the flywheel wins, and I believe this to be\ndue to its high efficiency.\n\nMy attempts at symbolic solution have been frustrating.\n\nI use either\n\ns" = (P/s\' - ( R + G(s) + c*(s\'+h)^2))/m or\ns" = (F/r - (R + G(s) + c*(s\'+h)^2)/m, with the first being a very difficult\ndifferential equation.\n\nit is on the face of F(s", s\', s) = 0 and should be solvable, but Mathcad won\'t\nhelp because P/s\' is divide by zero and I can\'t get the syntax to enter s\'>0 to\nwork. I may need a more recent version.\n\ns" = acceleration\nP = input Power\ns\' = speed\nR = Rolling resistance\nG(s) = w * d alt(s), that is, weight times change in USGS terrain altitude\nc = coefficient of drag expressed as G(s) at terminal speed and known location\ndivided by s\'^2 at that point, lbf/mph^2\nh = headwind, usually zero and for closed courses can be assumed to integrate\nto zero. For an oval track at constant speed, it does integrate to zero,\nalthough the effective c can change with wind direction in a way I haven\'t\nexplored very well yet.\nm = mass of rider, clothing, gear, tools, bicycle, water, and snacks\nF = Force applied to pedal\nr = ratio of (crank length times front cog teeth) to (wheel radius times rear\ncog teeth) (Gain ratio, I think it is called)\nw = m*g\nFt = total Force\nFt = ma\na = Ft/m\ns" = sigma F / m\n\nBy expressing alt(s) and G(s),\nwhich you must note is _not_ G(x) which eliminates the Pythagorean in the\ncomputations,\nas:\n\nsigma on i = 0 to 34 of Re(ab.i) * cos(k * i * s) + Im(ab.i) * sin( k * i * s)\n\nI find the opportunities for symbolic solution to be improved as this is an\nintegrable / differntiable expression.\n\nk = 2 * pi * Distance to be travelled\nRe(ab.0) = mean alititude\nIm(ab.0) = Distance\netc., accomplishing route specification with 34 lines (a page) of a Fourier\ndecomposition of alt(s) at about 500 points, providing a smoothed approximation\nto the jagged bitmap route profiles output by the route selection and profiling\nsoftware.\n\nIn a first order solver with dt = 1 sec, dr can = if (cadence high, -1, (if\n(cadence low, +1, otherwise 0)) but this has some trouble in the fourth order\nsolver. With r = 24/34 * (13/11) ^(Gear -1), Gear represents a cardinal number,\n1st, second gear, etc, and in the fourth order solver you get a column of Gear\nselections like 5.33, which operate well but do not predict what Gear the\nbicyclist will choose.\n\nThe gears on my bicycle are 24-35-51 / 34-28-23-19-16-13-11 so this is a good\napproximation. There are eleven distinct, widely spaced ratios, very evenly\nspaced, and ten duplicates. I use an R443 derailer and some shims on the chain\nrings, and a chain length / boom length adjuster used for multiple riders on my\nrecumbent will aid shifting since the rear derailer can\'t absorb that huge\ndifference in length. Although 24/11 and 51/34 are not good gears on my Peugeto\nmountain bike and aren\'t used since they score the chain, on the recumbent the\nchain line is very long and is held by two idlers, so all 21 gears are\navailable at any time without restriction. The adjuster hasn\'t been intalled\nyet so there are a few minor transmission problems. I stay in middle gear\nmostly.\n\nSo I more often just assume P constant, and P/s is just a force balance as if\nwe had an infinitely adjustable 100% efficient transmission. I adjust P as a\nboundary value with dP = 0 to solve for the Power required to traverse a given\nroute in time t2.\n\nThe syntax of the sbval expression used to solve the boundary value P for the\nscore (y2 - Distance) = 0 is\n\nsbval (y, t1, t2, load, score, d) where y is a vector of position, speed,\nacceleration, and power (the parameter), and d is the differential vector\nexpressing how each varies with dt and the other y.0, y.1, y.2, etc.\n\nI even wrote out the canonical solver with all twelve of the possible\nparameters in order to provide a menu from which, by deletion, any physics\nproblem in mechanics can be put in this differential solver\'s syntax. That\nhelped a lot.\n\nCan any of you solve\n\ns" = P/s\' - (R + cos(k*s) + c*s\'^2?\n\nor provide a few hints?\n\nReduction of order is not possible as far as I know for this equation, but t\ndoes not appear explicitly. s = s(t). s\' = ds/dt....\n\n\nYours,\n\nDoug Goncz ( ftp://users.aol.com/DGoncz/ )\n\nRead about my physics project at NVCC:\nhttp://groups.google.com/groups?q=dgoncz&scoring=d plus\n"bicycle", "fluorescent", "inverter", "flywheel", "ultracapacitor", etc.\nin the search box\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hey, gang.
I've completed fourth order Runge-Kutta simulation of electrically augmented,
flywheel augmented, and standard bicycles on USGS terrain routes in my
neighborhood. The results are that the flywheel wins, and I believe this to be
due to its high efficiency.
My attempts at symbolic solution have been frustrating.
I use either
s" = (P/s' - ( R + G(s) + c*(s'+h)^2))/m or
s" = (F/r - (R + G(s) + c*(s'+h)^2)/m, with the first being a very difficult
differential equation.
it is on the face of F(s", s', s) = and should be solvable, but Mathcad won't
help because P/s' is divide by zero and I can't get the syntax to enter s'>0 to
work. I may need a more recent version.
s" = acceleration
P = input Power
s' = speed
R = Rolling resistance
G(s) = w * d alt(s), that is, weight times change in USGS terrain altitude
c = coefficient of drag expressed as G(s) at terminal speed and known location
divided by s'^2 at that point, lbf/mph^2
h = headwind, usually zero and for closed courses can be assumed to integrate
to zero. For an oval track at constant speed, it does integrate to zero,
although the effective c can change with wind direction in a way I haven't
explored very well yet.
m = mass of rider, clothing, gear, tools, bicycle, water, and snacks
F = Force applied to pedal
r = ratio of (crank length times front cog teeth) to (wheel radius times rear
cog teeth) (Gain ratio, I think it is called)
w = m*gFt = total Force
Ft = maa = Ft/ms" = \sigma F / m
By expressing alt(s) and G(s),
which you must note is _not_ G(x) which eliminates the Pythagorean in the
computations,
as:
\sigma[/itex] on i = to 34 of Re(ab.i) * cos(k * i * s) + Im(ab.i) * sin( k * i * s)
I find the opportunities for symbolic solution to be improved as this is an
integrable / differntiable expression.
k = 2 * \pi * Distance to be travelled
Re(ab.0) = mean alititude
Im(ab.0) = Distance
etc., accomplishing route specification with 34 lines (a page) of a Fourier
decomposition of alt(s) at about 500 points, providing a smoothed approximation
to the jagged bitmap route profiles output by the route selection and profiling
software.
In a first order solver with dt = 1 sec, dr can = if (cadence high, -1, (if
(cadence low, +1, otherwise 0)) but this has some trouble in the fourth order
solver. With r = 24/34 * (13/11) ^(Gear -1), Gear represents a cardinal number,
1st, second gear, etc, and in the fourth order solver you get a column of Gear
selections like 5.33, which operate well but do not predict what Gear the
bicyclist will choose.
The gears on my bicycle are 24-35-51 / 34-28-23-19-16-13-11 so this is a good
approximation. There are eleven distinct, widely spaced ratios, very evenly
spaced, and ten duplicates. I use an R443 derailer and some shims on the chain
rings, and a chain length / boom length adjuster used for multiple riders on my
recumbent will aid shifting since the rear derailer can't absorb that huge
difference in length. Although 24/11 and 51/34 are not good gears on my Peugeto
mountain bike and aren't used since they score the chain, on the recumbent the
chain line is very long and is held by two idlers, so all 21 gears are
available at any time without restriction. The adjuster hasn't been intalled
yet so there are a few minor transmission problems. I stay in middle gear
mostly.
So I more often just assume P constant, and P/s is just a force balance as if
we had an infinitely adjustable 100% efficient transmission. I adjust P as a
boundary value with dP = to solve for the Power required to traverse a given
route in time t2.
The syntax of the sbval expression used to solve the boundary value P for the
score (y2 - Distance) = is
sbval (y, t1, t2, load, score, d) where y is a vector of position, speed,
acceleration, and power (the parameter), and d is the differential vector
expressing how each varies with dt and the other y.0, y.1, y.2, etc.
I even wrote out the canonical solver with all twelve of the possible
parameters in order to provide a menu from which, by deletion, any physics
problem in mechanics can be put in this differential solver's syntax. That
helped a lot.
Can any of you solve
s" = P/s' - (R + cos(k*s) + c*s'^2?
or provide a few hints?
Reduction of order is not possible as far as I know for this equation, but t
does not appear explicitly. s = s(t). [itex]s' = ds/dt....
Yours,
Doug Goncz ( ftp://users.aol.com/DGoncz/ )
Read about my physics project at NVCC:
http://groups.google.com/groups?q=dgoncz&scoring=d plus
"bicycle", "fluorescent", "inverter", "flywheel", "ultracapacitor", etc.
in the search box
I've completed fourth order Runge-Kutta simulation of electrically augmented,
flywheel augmented, and standard bicycles on USGS terrain routes in my
neighborhood. The results are that the flywheel wins, and I believe this to be
due to its high efficiency.
My attempts at symbolic solution have been frustrating.
I use either
s" = (P/s' - ( R + G(s) + c*(s'+h)^2))/m or
s" = (F/r - (R + G(s) + c*(s'+h)^2)/m, with the first being a very difficult
differential equation.
it is on the face of F(s", s', s) = and should be solvable, but Mathcad won't
help because P/s' is divide by zero and I can't get the syntax to enter s'>0 to
work. I may need a more recent version.
s" = acceleration
P = input Power
s' = speed
R = Rolling resistance
G(s) = w * d alt(s), that is, weight times change in USGS terrain altitude
c = coefficient of drag expressed as G(s) at terminal speed and known location
divided by s'^2 at that point, lbf/mph^2
h = headwind, usually zero and for closed courses can be assumed to integrate
to zero. For an oval track at constant speed, it does integrate to zero,
although the effective c can change with wind direction in a way I haven't
explored very well yet.
m = mass of rider, clothing, gear, tools, bicycle, water, and snacks
F = Force applied to pedal
r = ratio of (crank length times front cog teeth) to (wheel radius times rear
cog teeth) (Gain ratio, I think it is called)
w = m*gFt = total Force
Ft = maa = Ft/ms" = \sigma F / m
By expressing alt(s) and G(s),
which you must note is _not_ G(x) which eliminates the Pythagorean in the
computations,
as:
\sigma[/itex] on i = to 34 of Re(ab.i) * cos(k * i * s) + Im(ab.i) * sin( k * i * s)
I find the opportunities for symbolic solution to be improved as this is an
integrable / differntiable expression.
k = 2 * \pi * Distance to be travelled
Re(ab.0) = mean alititude
Im(ab.0) = Distance
etc., accomplishing route specification with 34 lines (a page) of a Fourier
decomposition of alt(s) at about 500 points, providing a smoothed approximation
to the jagged bitmap route profiles output by the route selection and profiling
software.
In a first order solver with dt = 1 sec, dr can = if (cadence high, -1, (if
(cadence low, +1, otherwise 0)) but this has some trouble in the fourth order
solver. With r = 24/34 * (13/11) ^(Gear -1), Gear represents a cardinal number,
1st, second gear, etc, and in the fourth order solver you get a column of Gear
selections like 5.33, which operate well but do not predict what Gear the
bicyclist will choose.
The gears on my bicycle are 24-35-51 / 34-28-23-19-16-13-11 so this is a good
approximation. There are eleven distinct, widely spaced ratios, very evenly
spaced, and ten duplicates. I use an R443 derailer and some shims on the chain
rings, and a chain length / boom length adjuster used for multiple riders on my
recumbent will aid shifting since the rear derailer can't absorb that huge
difference in length. Although 24/11 and 51/34 are not good gears on my Peugeto
mountain bike and aren't used since they score the chain, on the recumbent the
chain line is very long and is held by two idlers, so all 21 gears are
available at any time without restriction. The adjuster hasn't been intalled
yet so there are a few minor transmission problems. I stay in middle gear
mostly.
So I more often just assume P constant, and P/s is just a force balance as if
we had an infinitely adjustable 100% efficient transmission. I adjust P as a
boundary value with dP = to solve for the Power required to traverse a given
route in time t2.
The syntax of the sbval expression used to solve the boundary value P for the
score (y2 - Distance) = is
sbval (y, t1, t2, load, score, d) where y is a vector of position, speed,
acceleration, and power (the parameter), and d is the differential vector
expressing how each varies with dt and the other y.0, y.1, y.2, etc.
I even wrote out the canonical solver with all twelve of the possible
parameters in order to provide a menu from which, by deletion, any physics
problem in mechanics can be put in this differential solver's syntax. That
helped a lot.
Can any of you solve
s" = P/s' - (R + cos(k*s) + c*s'^2?
or provide a few hints?
Reduction of order is not possible as far as I know for this equation, but t
does not appear explicitly. s = s(t). [itex]s' = ds/dt....
Yours,
Doug Goncz ( ftp://users.aol.com/DGoncz/ )
Read about my physics project at NVCC:
http://groups.google.com/groups?q=dgoncz&scoring=d plus
"bicycle", "fluorescent", "inverter", "flywheel", "ultracapacitor", etc.
in the search box